|4x + 3| = 9 + 2x

Since the variable is on both sides of the equation, you would, at the end, check for extraneous solutions.

Extraneous solutions are solutions that do not work with the equation, therefore they are "extra" solutions and un-included in your final answer.

Start the problem by splitting the equation into two equations, a positive case and a negative case. Your two equations would look like:

4x + 3 = 9 + 2x {positive case}4x + 3 = -(9 + 2x) {negative case}---Solving the equations---[POSITIVE CASE]

Let's solve for the positive case first. Start by subtracting 3 from both sides of the equation.

4x + 3 = 9 + 2x becomes 4x = 6 + 2x

Now subtract 2x from both sides of the equation.

4x = 6 + 2x becomes 2x = 6

Finish off the problem by dividing both sides by 2 to isolate the variable x.

2x = 6 becomes x = 3.---[NEGATIVE CASE]

Now let's solve for x in the negative case. Start by distributing the negative sign (-) inside the parentheses.

4x + 3 = -(9 + 2x) becomes 4x + 3 = -9 - 2x

Subtract 3 from both sides just like the positive case.

4x + 3 = -9 - 2x becomes 4x = -12 - 2x

Now add 2x to both sides of the equation.

4x = -12 - 2x becomes 6x = -12

Finish off the problem by dividing both sides by 6 to isolate the variable x.

6x = -12 becomes x = -2.---Checking for extraneous solutions---[CHECKING X = 3]

To check for extraneous solutions, or solutions that do not work, substitute what you got for x back into the original absolute value equation: |4x + 3| = 9 + 2x. Substitute 3 and -2 into the equation. Let's start by substituting 3 for x.

|4x + 3| = 9 + 2x becomes |4(3) + 3| = 9 + 2(3)

Start by multiplying 4 and 3 together inside the absolute value symbols.

|4(3) + 3| = 9 + 2(3) becomes |(12) + 3| = 9 + 2(3)

Now multiply 2 and 3 together.

|(12) + 3| = 9 + 2(3) becomes |(12) + 3| = 9 + (6)

Add 12 and 3 together inside the absolute value symbols; also add 9 and 6 together.

|(12) + 3| = 9 + (6) becomes |(15)| = (15), which is the same as 15 = 15.

15 = 15 is a true statement so this means that 3 is a solution to the absolute value equation, so it is not an extraneous solution.

---[CHECKING X = -2]

Let's see if -2 is a solution or not - substitute -2 for x into the equation: |4x + 3| = 9 + 2x.

|4x + 3| = 9 + 2x becomes |4(-2) + 3| = 9 + 2(-2)

Multiply 4 and -2 inside the absolute value symbols.

|4(-2) + 3| = 9 + 2(-2) becomes |(-8) + 3| = 9 + 2(-2)

Multiply 2 and -2.

|(-8) + 3| = 9 + 2(-2) becomes |(-8) + 3| = 9 + (-4)

Add -8 and 3 inside the absolute value symbols; also add 9 and -4.

|(-8) + 3| = 9 + (-4) becomes |(-5)| = (5), which is the same as 5 = 5.

5 = 5 is a true statement so that means it is not an extraneous solution. After checking for extraneous solutions, we have come to the conclusion that the two answers for the equation --> I4x + 3I = 9 + 2x <-- are x = 3 or x = 2.